To solve the system of equations:

$3x + y = 13$ $13x - 2y = -5$

we can use either substitution or elimination. Let's go through it step-by-step using the elimination method.

Step 1: Set up the Equations

1. $3x + y = 13$
2. $13x - 2y = -5$

Step 2: Make the Coefficients of $y$ the Same

To eliminate $y$, we want the coefficients of $y$ to be equal in both equations. We can do this by multiplying the first equation by $2$ to match the $2y$ term in the second equation.

$2 \times (3x + y = 13) \Rightarrow 6x + 2y = 26$

Now we have:

1. $6x + 2y = 26$
2. $13x - 2y = -5$

Step 3: Add the Equations

Now we can add both equations to eliminate $y$:

$(6x + 2y) + (13x - 2y) = 26 + (-5)$

Simplifying this, we get:

$19x = 21$

So,

$x = \frac{21}{19}$

Step 4: Substitute $x$ Back to Find $y$

Now, substitute $x = \frac{21}{19}$ back into the first equation:

$3\left(\frac{21}{19}\right) + y = 13$

$\frac{63}{19} + y = 13$

Convert $13$ to a fraction with a denominator of $19$:

$\frac{63}{19} + y = \frac{247}{19}$

Subtract $\frac{63}{19}$ from both sides:

$y = \frac{247 - 63}{19} = \frac{184}{19}$

Solution

The solution to the system is:

$x = \frac{21}{19}, \quad y = \frac{184}{19}$

Let me know if you would like further details, or if you have any questions!

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Here are five related questions for further exploration:

1. How can we solve the same system using substitution?
2. What would be the solution if we changed the constants in the equations?
3. How can we verify the solution by substituting back into both original equations?
4. How does elimination work when both variables have non-matching coefficients in both equations?
5. What are other methods for solving systems of equations beyond elimination and substitution?

Tip: Always check your solution by substituting back into the original equations to ensure accuracy.